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Finger [1]
3 years ago
8

On a coordinate plane, 2 straight lines are shown. The first solid line has a negative slope and goes through (negative 2, 3) an

d (0, negative 1). Everything to the left of the line is shaded. The second dashed line has a negative slope and goes through (0, 2) and (1, 0). Everything to the right of the line is shaded.
Which inequality pairs with y≤−2x−1 to complete the system of linear inequalities represented by the graph?

y −2x+2
y 2x−2
Mathematics
2 answers:
Firlakuza [10]3 years ago
9 0

Answer:

C on edge

Step-by-step explanation:

dimulka [17.4K]3 years ago
3 0

Answer:B  i think......

Step-by-step explanation:

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Try to sketch by hand the curve of intersection of the parabolic cylinder y = x2 and the top half of the ellipsoid x2 + 7y2 + 7z
vovikov84 [41]

Plug y=x^2 into the equation of the ellipsoid:

x^2+7(x^2)^2+7z^2=49

Complete the square:

7x^4+x^2=7\left(x^4+\dfrac{x^2}7+\dfrac1{14^2}-\dfrac1{14^2}\right)=7\left(x^2+\dfrac1{14}\right)^2+\dfrac1{28}

Then the intersection is such that

7\left(x^2+\dfrac1{14}\right)^2+7z^2=\dfrac{1371}{28}

\left(x^2+\dfrac1{14}\right)^2+z^2=\dfrac{1371}{196}

which resembles the equation of a circle, and suggests a parameterization is polar-like coordinates. Let

x(t)^2+\dfrac1{14}=\sqrt{\dfrac{1371}{196}}\cos t\implies x(t)=\pm\sqrt{\sqrt{\dfrac{1371}{196}}\cos t-\dfrac1{14}}

y(t)=x(t)^2=\sqrt{\dfrac{1371}{196}}\cos t-\dfrac1{14}

z=\sqrt{\dfrac{1371}{196}}\sin t

(Attached is a plot of the two surfaces and the intersection; red for the positive root x(t), blue for the negative)

4 0
3 years ago
Order -6 1/4, -6.35, -6 1/5, and 6.1 from greatest to least. Explain.
san4es73 [151]
If you turn all the numbers into decimals it becomes, -6.25, -6.35, -6.20, and, 6.1 so in order it would go 6.1, -6.20, -6.25, -6.35
3 0
3 years ago
My sister needs help!
arlik [135]

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6 0
3 years ago
Read 2 more answers
a student has reflected the patterns as given below are the reflections correct and what advice would yiu give them
siniylev [52]

Answer:

Step-by-step explanation:

We need the rest of the question please.

8 0
3 years ago
The coordinates of the point that is a reflection of y(-4-2) across the x-axis are ( ?
Darya [45]

Assuming your point is Y(-4, -2), its reflection across y=0 simply changes the sign of the y-coordinate.

The point after reflection is Y'(-4, 2).

7 0
3 years ago
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